|
|
SECTION 29.3 • Torque on a Current Loop in a Uniform Magnetic Field 905 Now suppose that the uniform magnetic field makes an angle ! * 90° with a line perpendicular to the plane of the loop, as in Figure 29.14. For convenience, we assume B is perpendicular to sides # and $. In this case, the magnetic forces F 1 and F 3 exerted on sides ! and " cancel each other and produce no torque because they pass F 2 and F 4 acting on sides # and $ produce a torque about any point. Referring to the end view shown in Figure F 2 about the point O is equal to (b/2) sin !. Likewise, the moment arm of F 4 about O is also (b/2) sin !. Because F 2 " F 4 " IaB, the magnitude of the net torque about O is where A " ab is the area of the loop. This result shows that the torque has its A convenient expression for the torque exerted on a loop placed in a uniform magnetic field B is (29.9) where A, the vector shown in Figure 29.14, is perpendicular to the plane of the loop and has a magnitude equal to the area of the loop. We determine the direction of A using the right-hand rule described in Figure 29.15. When you curl the fingers of your right hand A. As we see in Figure 29.14, the loop tends to rotate in the direction of decreasing values of ! A rotates toward the direction of the magnetic field). The product I A is defined to be the magnetic dipole moment " (often simply called the “magnetic moment”) of the loop: (29.10) The SI unit of magnetic dipole moment is ampere-meter 2 (A # m 2 ). Using this definition, we can express the torque exerted on a current-carrying loop in a magnetic field B as (29.11) Note that this result is analogous to Equation 26.18, # " p ! E, for the torque exerted on an electric dipole in the presence of an electric field E, where p is the electric dipole moment. # $ " ! B " " I
A # " I
A ! B " IAB sin ! " IaB # b 2 sin ! $ ' IaB # b 2 sin ! $ " IabB sin ! ) " F 2
b 2 sin ! ' F 4
b 2 sin ! F 2 F 4 O B A b 2 – sin θ b 2 – θ θ θ # $ × Active Figure 29.14 An end view of the loop in Figure 29.13b rotated through an angle with respect to the magnetic field. If B is at an angle ! with respect to vector A, which is perpendicular to the plane of the loop, the torque is IAB sin ! where the magnitude of A is A, the area of the loop. At the Active Figures link at http://www.pse6.com, you can choose the current in the loop, the magnetic field, and the initial orientation of the loop and observe the subsequent motion. Torque on a current loop in a magnetic field Magnetic dipole moment of a current loop Torque on a magnetic moment in a magnetic field A I µ Figure 29.15 Right-hand rule for determining the direction of the vector A. The direction of the magnetic moment " is the same as the direction of A. |